Gọi \(d=\left(18a+2;30a+3\right)\)
\(\Rightarrow\hept{\begin{cases}18a+2⋮d\\30a+3⋮d\end{cases}}\)
\(\Rightarrow5\left(18a+2\right)-3\left(30a+3\right)⋮d\)
\(\Rightarrow\left(90a+10\right)-\left(90a+9\right)⋮d\)
\(\Rightarrow1⋮d\)
\(\Rightarrow d=1\)
Vậy \(\left(18a+2;30a+3\right)=1\).
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